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Dear Abhi,
you may ask the question that you are not able to solve. we will give you full explanation about that.

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Vaibhav

6/10/2013 03:41:38 pm

Q.3 has three parts and the answer to th third part is not given

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shubham

6/15/2013 03:21:08 am

thermodynamic hc verma Qno-21.

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theyatin

6/15/2013 05:33:35 pm

dear.
part a) as wall can not be moved thus change in internal energy =0 also change in total charge =0
as internal energy is zero thus work done is also zero.
part b)
as we know PV/RT=constant
thus we can compare this equation for both temperatures.
later putting value of T we have (p1+p2)*T2/lembda
now putting this in equation we have the answer.

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ashlesha

11/18/2013 07:24:44 pm

pooja

7/4/2013 01:56:04 am

1 mole of agas expands with temperature Tsuch that its volume, V=kT^2 ,where k is a constant.if the temp. Of thegas changes by 60 degree C than find the work done by the gas ?

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raghav

5/4/2015 04:10:45 am

V=kT^2
=>V/T^2=k
V/(VP)^2=k

=>V(P)^2 = const

W={integration}(P dV)

I guess you will be able to solve from here on 😊

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pooja

7/4/2013 02:04:22 am

A cylinder fitted with a piston contain an ideal monoatomic gas at a temp.of 400k. The piston is held fixed with heat del Q is given to the gas ,it is found that the temprature of the gas has increased by 20 k . In a isobaric processthe same heat is supplied slowly to it.find the change in temperature in the second process?

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pooja

7/4/2013 02:06:16 am

Sir please explain me Q - 21( 2) part.

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Aymanzoor

12/3/2013 12:10:08 am

SIr please explain Q#3 part c
how is it 400J

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tazeen

1/7/2014 02:35:18 am

in question. 5 to calculate work done why we take 1/2 pv

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theyatin

1/8/2014 02:17:16 am

dear,
work done=F*d
P=F/A
F=P*A
W.D=P*A*d=P*volume

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ketan

10/13/2014 12:02:52 am

why 1/2 pv??

Study

1/22/2014 02:51:00 am

In Q7 for process AB to calclate ar of ABC why do we take the base(AC) as (30+10)
Shouldnt it be (30-10) ???

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theyatin

1/23/2014 01:02:45 am

dear
its typing mistake
check calculations its 30-10 n right

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Study

3/30/2014 01:49:02 am

No sir....if we take 30-10 we will get diff answer..
Nd dis is same for all questions

theyatin

4/1/2014 04:31:43 am

dear,
a typing mistake may lead to wrong answer too.
so just focus on right method and try by yourself if method is right you ll get right answer.
and you are right we get other answer if we tak 30-10

sir,v stands for volume,k is a constant,T stands for temperature,pls find out the answer

Mou

2/11/2014 03:13:19 pm

Sir Plz explain q number 10

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theyatin

2/13/2014 11:48:50 pm

dear,
short answer Q or exercise Q.

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Mou

2/16/2014 03:38:29 am

Sir exercise q 10

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theyatin

2/19/2014 01:34:23 am

dear,
as PV=work done
thus from graph area of circle is P*V
thus area of circle=pi*r*r

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Shrestha

2/20/2014 03:40:39 pm

Sir please explain hc verma thermodynamics exercise q no.20

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ER.MIRJAIK

5/9/2014 02:05:01 pm

dear shrestha
one thing should be clear that internal energy is the function of temperature means as temperature increases internal energy also increases and intermolecular energy is the function of volume as intermolecu;ar energy increases volume also increases.
here in this problem we just using the formula for the change in internal energy as per first law of thermodynamics one thing it should be clear that pressure used is atmospheric pressure and dQ is the net heat change through whole process.as u know density of water is 1000kg/m^3 and steam density is given to you then delta is only the change in that.
here dw= p* delta v is the expansion work

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seema

4/25/2014 06:28:10 am

sir please explain ques no 18

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ER. MIRJAIK

5/9/2014 02:17:21 pm

dear seema
as it is isochoric process net change in volume is zero so no change in internal energy is going to take place hence dw=dq
but change in work done is to calculated through cyclic process hence we use gas equation.put values in the formula and get the answer

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aditya

5/7/2014 10:33:10 pm

sir ,please explain question number -21 and 22 of thernodyanamics of hc verma

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ER.MIRJAIK

5/9/2014 01:37:47 pm

Q.N21
dear.
part a) as wall can not be moved thus change in internal energy =0 also change in total charge =0
as internal energy is zero thus work done is also zero.
part b)
as we know PV/RT=constant
thus we can compare this equation for both temperatures.
later putting value of T we have (p1+p2)*T2/lembda
now putting this in equation we have the answer.

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Admin

6/22/2014 01:50:58 pm

Dear Students,
Our Teacher is unavailable for few days. We are sorry for this inconvenience. Our teacher will give your answers soon.

go on this website
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it has all the answers and better.

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ashutosh m. reddy (dps secun'bad where principal considers us as srcs of money!!!)

6/23/2014 01:53:43 am

can u tell me wat is lambda in Q21 nd frm were d frmula of t was was derived wch is p1p2......../lambda????

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Admin

6/23/2014 01:50:34 pm

sorry ashutosh v cant answer ur Q as it is very difficult .... but u can post d same Q on odr sites 2 get ur aanswer.......so v r sorry ......v jst keep dis site 2 earn money thru ads....nd it is only for dumb students....